Solve 3+i/2+sqrt{2i}

$$\frac { 3 + i } { 2 + \sqrt { 2 } i }$$

Evaluate

$$\sqrt{2}\left(\frac{1}{6}-\frac{1}{2}i\right)+\left(1+\frac{1}{3}i\right)\approx 1.23570226-0.373773448i$$

Solution Steps

Rationalize the denominator of \frac{3+i}{2+i\sqrt{2}} by multiplying numerator and denominator by 2-i\sqrt{2}.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{\left(2+i\sqrt{2}\right)\left(2-i\sqrt{2}\right)}$$

Consider \left(2+i\sqrt{2}\right)\left(2-i\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{2^{2}-\left(i\sqrt{2}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{4-\left(i\sqrt{2}\right)^{2}}$$

Expand \left(i\sqrt{2}\right)^{2}.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{4-i^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate i to the power of 2 and get -1.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{4-\left(-\left(\sqrt{2}\right)^{2}\right)}$$

The square of \sqrt{2} is 2.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{4-\left(-2\right)}$$

Multiply -1 and -2 to get 2.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{4+2}$$

Add 4 and 2 to get 6.

$$\frac{\left(3+i\right)\left(2-i\sqrt{2}\right)}{6}$$

Divide \left(3+i\right)\left(2-i\sqrt{2}\right) by 6 to get \left(\frac{1}{2}+\frac{1}{6}i\right)\left(2-i\sqrt{2}\right).

$$\left(\frac{1}{2}+\frac{1}{6}i\right)\left(2-i\sqrt{2}\right)$$

Use the distributive property to multiply \frac{1}{2}+\frac{1}{6}i by 2-i\sqrt{2}.

$$1+\frac{1}{3}i+\left(\frac{1}{6}-\frac{1}{2}i\right)\sqrt{2}$$

Real Part

$$\frac{\sqrt{2}}{6} + 1 = 1.23570226$$